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In a bonus exercise last year, we were asked to compute the completion in general of such a stalk on a smooth manifold of dimension $n$ (it is isomorphic to the ring of formal power series over $\mathbb{R}$ in $n$ unknowns). It's clear that this is a bad case to work with, since smooth manifolds admit bump functions (which allows us to prove that there exists a nonzero element in the intersection of all (finite) powers of the maximal ideal), and therefore the completion contains very little data.

However, what kind of "stuff" does this technique allow one to do in the analytic/holomorphic cases? Similarly, in the algebraic case, we can often look at the henselization for the same information, but I still am not really sure why one would want to do so in the first place. That is, what geometric idea corresponds to the idea of "completion" (including henselization and strict henselization depending on the context) in the same way that localization at a prime corresponds to taking stalks geometrically?

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  • $\begingroup$ From what I can gather, it seems that the henselization has better finiteness properties than the completion but contains all of the important data in many important cases (It is my understanding that this is the geometric content of the Artin approximation theorem and the result of Popescu on General Néron Desingularization) $\endgroup$ Sep 12, 2010 at 9:33
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    $\begingroup$ Dear Harry: Do you know about excellent rings (Chapter 13 of Matsumura's "Commutative Algebra" -- not "Commutative Ring Theory" -- and section 7.7 or 7.8 in EGA IV$_3$)? Gives powerful technique to prove for "most" (not all!) local noetherian rings that properties hold if and only if they do for completion. As Torsten notes, if $X$ is an algebraic scheme over $\mathbf{C}$ and $x \in X(\mathbf{C})$ then $O_ {X,x}$ and $O_ {X^{\rm{an}},x}$ have the same completion. So $X$ is reduced, normal, Cohen-Macaulay, etc. iff $X^{\rm{an}}$ is. I doubt there's a proof without going through completion. $\endgroup$
    – BCnrd
    Sep 12, 2010 at 13:26
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    $\begingroup$ Dear Harry: Completions admit technique of working by successive approx., not available with henselization (until Artin approx. came along, but that often doesn't suffice). Learn deformation theory. Henselization and strict henselization are the "local rings" for the etale topology, and so are ubiquitous for local calculations on alg. spaces and DM stacks; local rings for their Zar. topologies are often useless. As Torsten notes, solving problem over str. hens. often allows to descend to actual etale nbhd; can't do with completion. Read Knutson's "Alg. Spaces" and sec. 3.6 of "Neron Models". $\endgroup$
    – BCnrd
    Sep 12, 2010 at 13:37
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    $\begingroup$ Dear Harry: Here is a geometric example showing merits of completion and henselization. Let $X$ be proper flat scheme over local noetherian $R$ with fiber dim. 1. Is it projective? Using ample line bundle on special fiber, infinitesimal def. theory and formal GAGA (for coherent sheaves) imply lifts to $X' := X_ {\widehat{R}}$, so $X'$ projective over $\widehat{R}$. If $R$ is excellent, Artin approx. implies such a lift then exists on $X'' := X_ {R^{\rm{h}}}$, so $X''$ is projective over $R^{\rm{h}}$. Thus, $X$ is projective over a (residually trivial) etale nbhd of $R$. Seems best can do. $\endgroup$
    – BCnrd
    Sep 12, 2010 at 13:45
  • $\begingroup$ Dear BCnrd: Thanks, I will check it out. Also, yes, I am familiar with excellent rings. $\endgroup$ Sep 12, 2010 at 18:18

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First I think you are a little bit unfair when you say that the completion of the ring of germs of $C^\infty$-functions "contains very little". Mapping a function into that completion gives you the Taylor series of function which contains a lot of data about the function even though you certainly miss a significant amount of information...

Anyway, you are certainly right looking at the (strict) Henselisation of a local ring is often better than looking at the completion. This is mainly because the Henselisation is a direct limit of very well-behaved extension rings and hence many things which are defined over the Henselisation will be defined over one of these extension rings. However, if you look at the local rings of analytic spaces they already are Henselian yet you still use the completion even in those case. The reason for that is that constructing elements of the completion can be done by a step by step by step procedure, constructing one term of a Taylor expansion at the time. In classical complex analysis one then usually performs a closer analysis and shows that the resulting power series is actually convergent but the first step is still important.

From that point of view the Artin approximation (and generalisations) gives a very general criterion for when certain processes automatically give convergent power series provided that it gives any power series at all. Note that there are many more classical results which allow you to pass to the completion without using something like the approximation theorem. One such is that the completion (of a Noetherian local ring) gives a faithfully flat extensions which in particular allows you to check many equalities by passing to the completion.

However, if we back down to an algebraic scheme over $\mathbb C$ then you get several local rings; the local ring of a point, its Henselisation, the ring of germs of analytic functions and the completion. The ring of germs of analytic functions doesn't make algebraic sense so Henselisation and completion are algebraic substitutes (from "both sides"). The Henselisation is closer to the original ring which is an advantage but is often more difficulat to work with, while the completion is easier to work with but it is more difficult to get back to the original ring. The approximation theorems should be seen as a very powerful way of getting back to the Henselisation and then one can work at getting from the Henselisation to the original or one can stay at the Henselisation (this is what the étale topology is all about). (This is the optimistic view on the approximation theorems, rather than the pessimistic one that they say that you never need to pass to the completion as everything already lies in the Henselisation.)

Added: To add some specific examples. On is the proof that a regular local ring is a UFD (let us assume that it contains a copy of its residue field to simplify). This one shows by first showing that the ring is UFD it its completion is. For the completion which then is a power series ring over the residue field one can use the Weierstrass preparation theorem to show that one can reduce to the polynomial ring in the last variable over the power series ring in all but the last. This is a UFD by induction and the fact that the polynomial ring over a UFD is a UFD ("Gauss lemma"). Here the Henselisation does not appear at all.

Another (far more sophisticated) example is the one which I guess was one of Artin's motivation for the approximation theorem to begin with. Here one wants to show that some functor is representable by an algebraic space. This means constructing a universal element over some suitable base. The first step is to use deformation theory to show that the functor (for a fixed point over some field) is prorepresentable. This is done exactly by showing that it is representable over the category of local Artinian rings for which a fixed power of the maximal ideal is zero. This is done by induction over the fixed power (and hence can be said to do a Taylor expansion one going from one order to the next). The end result (if everything works!) is a formal deformation over some complete local ring. Then one uses further properties of the functor to show that this formal deformation is given by an actual element of the functor (this typically uses Grothendieck's GAGA-type results for formal schemes). Then one uses the approximation theorem to show that the element comes from the Henselisation of something of finite type (there is an extra trickyness in that the complete ring is not known beforehand to descend as some such Henselisation). One then stops there and uses the universality to get gluing data for an algebraic space.

Note that there are situations when this doesn't work. A particular class of examples arise when one is dealing with differential equations: There are differential equations (with coefficients in the local ring of a smooth variety) which has no solution in the Henselisation but does have a solution in germs of analytic functions and there are differential equations that have formal power series solutions but no convergent ones.

As for this setup for a general locally ringed space I do not know of any general results except for the situation where a problem can be reduced to a commutative algebra type problem concerning the local rings of the space which then can be solved by forgetting about the space altogether.

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  • $\begingroup$ Dear Torsten, while this is certainly informative and a very good answer, could you touch a little bit more on what kinds of geometric data are available by looking at the completion/henselization? $\endgroup$ Sep 12, 2010 at 11:24
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    $\begingroup$ Torsten's answer leaves aside a related question I had never thought about: Assume $X$ and $Y$ are schemes of finite type over a field $k$, and $x\in X$, $y\in Y$ are points such that the completions are isomorphic. Are the henselizations isomorphic? $\endgroup$ Sep 12, 2010 at 14:41
  • $\begingroup$ Probably yes, by Artin approximation applied to a morphism $\widehat{X}\to Y$ inducing an isomorphism on completions. $\endgroup$ Sep 12, 2010 at 15:03
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    $\begingroup$ Dear Laurent: the affirmative answer in the relative case of points in the special fiber of a finite type scheme over an excellent local ring (i.e., combining Artin approximation with a "formal isomorphism" to make an isomorphism between henselizations and then between etale neighborhoods) is used in the proof of the etale-local structure of ordinary double pts (defining "ordinary double pt" via condition on completed local rings on geom. fibers, say). In the example you mention that relates to the special case $x$ and $y$ closed, even though the argument works in general. $\endgroup$
    – BCnrd
    Sep 12, 2010 at 15:26

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